Two charges, Q1= 2.70 µC, and Q2= 5.20 µC are located at points (0,-2.00 cm ) an

Question

Two charges, Q1= 2.70 µC, and Q2= 5.20 µC are located at points (0,-2.00 cm ) and (0,+2.00 cm), as shown in the figure.

(The figure is like a sideways T, almost like a plus sign without the left side of the plus. Point P is located on the right side of the x axis at the end. Point Q1 is on the y axis at the bottom end, and point Q2 is located at the top end of the y axis).

a) What is the magnitude of the electric field at point P, located at (5.50 cm, 0), due to Q1 alone?

b) What is the x-component of the total electric field at P?

c) What is the y-component of the total electric field at P?

d) What is the magnitude of the total electric field at P?

e) Now let Q2 = Q1 = 2.70 µC. Note that the problem now has a symmetry that you should exploit in your solution. What is the magnitude of the total electric field at P?

f) Given the symmetric situation of the previous problem, what is the magnitude of the force on an electron placed at point P?

Please show all work! I will rate really good! Thanks! (:

Explanation / Answer

r 2= (22+5.52) = 34.25 cm2 = 34.25 x 10-4 m2    or    r =34.25 = 5.85 cm

A) Electric field at P by Q1 alone as E1 = k.Q1/r2 = 9x109.(2.70x10-6)/34.25 x10-4

                                                    E1 = 7.095 x 106 N/C

B) from right triangle we know that cos = 5.50/5.85 = 0.94

                                                    sin = 2/5.85 = 0.34

Electric field at P by Q2 alone as E2 = k.Q2/r2 = 9x109.(5.20x10-6)/(34.25x10-4)

                                                 E2 = 13.664 x 106 N/C

X componen as Ex = E1 cos + E2 cos = 7.095x106.(0.94) + 13.664x106.(0.94) = 19.513 x106 N/C

C) Y component as Ey = E1 sin - E2 sin = 7.095x106(0.34) - 13.644x106.(0.34) = -2.227 x 106 N/C

D) Etot = (Ex2 + Ey2) = [(19.513x106)2+(-2.227x106)2] =19.64 x106 N/C

E) If Q1 =Q2 = 2.70C that E1 = E2 and the y component of E1 and E2 cancel out so Etot = 2E1 cos

     Etot = 2.(7.095x106).(0.94) = 13.339 x 106 N/C

F) Electric force F =q.Etot = (1.6x10-19).(13.339x106) = 2.134 x10-12 N

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